Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(f(a, b), x) → f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(f(a, b), x) → f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(f(a, b), x) → F(c, f(b, f(a, x)))
F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(x, y)
F(f(a, b), x) → F(b, f(a, f(c, f(b, f(a, x)))))
F(f(a, b), x) → F(b, f(a, x))
F(f(a, b), x) → F(a, f(c, f(b, f(a, x))))
F(x, f(y, z)) → F(f(x, y), z)

The TRS R consists of the following rules:

f(f(a, b), x) → f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

F(f(a, b), x) → F(c, f(b, f(a, x)))
F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(x, y)
F(f(a, b), x) → F(b, f(a, f(c, f(b, f(a, x)))))
F(f(a, b), x) → F(b, f(a, x))
F(f(a, b), x) → F(a, f(c, f(b, f(a, x))))
F(x, f(y, z)) → F(f(x, y), z)

The TRS R consists of the following rules:

f(f(a, b), x) → f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(f(a, b), x) → F(c, f(b, f(a, x)))
F(f(a, b), x) → F(b, f(a, f(c, f(b, f(a, x)))))
F(f(a, b), x) → F(b, f(a, x))
The remaining pairs can at least be oriented weakly.

F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(x, y)
F(f(a, b), x) → F(a, f(c, f(b, f(a, x))))
F(x, f(y, z)) → F(f(x, y), z)
Used ordering: Polynomial interpretation [25]:

POL(F(x1, x2)) = x1   
POL(a) = 1   
POL(b) = 0   
POL(c) = 0   
POL(f(x1, x2)) = x1   

The following usable rules [17] were oriented:

f(x, f(y, z)) → f(f(x, y), z)
f(f(a, b), x) → f(b, f(a, f(c, f(b, f(a, x)))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(x, y)
F(f(a, b), x) → F(a, f(c, f(b, f(a, x))))
F(x, f(y, z)) → F(f(x, y), z)

The TRS R consists of the following rules:

f(f(a, b), x) → f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.